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  • LeetCode 176. Second Highest Salary

    https://leetcode.com/problems/second-highest-salary/description/

    Write a SQL query to get the second highest salary from the Employee table.

    +----+--------+
    | Id | Salary |
    +----+--------+
    | 1  | 100    |
    | 2  | 200    |
    | 3  | 300    |
    +----+--------+
    

    For example, given the above Employee table, the query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.

    +---------------------+
    | SecondHighestSalary |
    +---------------------+
    | 200                 |
    +---------------------+

    Solution

    Approach: Using sub-query and LIMIT clause [Accepted]

    Algorithm

    Sort the distinct salary in descend order and then utilize the LIMIT clause to get the second highest salary.

    SELECT DISTINCT
        Salary AS SecondHighestSalary
    FROM
        Employee
    ORDER BY Salary DESC
    LIMIT 1 OFFSET 1
    

    However, this solution will be judged as 'Wrong Answer' if there is no such second highest salary since there might be only one record in this table. To overcome this issue, we can take this as a temp table.

    MySQL

    SELECT
        (SELECT DISTINCT
                Salary
            FROM
                Employee
            ORDER BY Salary DESC
            LIMIT 1 OFFSET 1) AS SecondHighestSalary
    ;
    

     The LIMIT clause can be used to constrain the number of rows returned by the SELECT statement. LIMIT takes one or two numeric arguments, which must both be nonnegative integer constants, with these exceptions:

    • Within prepared statements, LIMIT parameters can be specified using ? placeholder markers.

    • Within stored programs, LIMIT parameters can be specified using integer-valued routine parameters or local variables.

    With two arguments, the first argument specifies the offset of the first row to return, and the second specifies the maximum number of rows to return. The offset of the initial row is 0 (not 1):

    SELECT * FROM tbl LIMIT 5,10;  # Retrieve rows 6-15

    To retrieve all rows from a certain offset up to the end of the result set, you can use some large number for the second parameter. This statement retrieves all rows from the 96th row to the last:

    SELECT * FROM tbl LIMIT 95,18446744073709551615;

    With one argument, the value specifies the number of rows to return from the beginning of the result set:

    SELECT * FROM tbl LIMIT 5;     # Retrieve first 5 rows

    In other words, LIMIT row_count is equivalent to LIMIT 0, row_count.

    For prepared statements, you can use placeholders. The following statements will return one row from the tbl table:

    SET @a=1;
    PREPARE STMT FROM 'SELECT * FROM tbl LIMIT ?';
    EXECUTE STMT USING @a;

    The following statements will return the second to sixth row from the tbl table:

    SET @skip=1; SET @numrows=5;
    PREPARE STMT FROM 'SELECT * FROM tbl LIMIT ?, ?';
    EXECUTE STMT USING @skip, @numrows;

    For compatibility with PostgreSQL, MySQL also supports the LIMIT row_count OFFSET offset syntax.

    If LIMIT occurs within a subquery and also is applied in the outer query, the outermost LIMIT takes precedence. For example, the following statement produces two rows, not one:

     
    (SELECT ... LIMIT 1) LIMIT 2;

     

     1 Create table If Not Exists Employee (Id int, Salary int);
     2 Truncate table Employee;
     3 insert into Employee (Id, Salary) values ('1', '100');
     4 insert into Employee (Id, Salary) values ('2', '200');
     5 insert into Employee (Id, Salary) values ('3', '300');
     6 
     7 # Write your MySQL query statement below
     8 SELECT
     9   (SELECT DISTINCT Salary
    10   FROM Employee 
    11   ORDER BY Salary DESC
    12   LIMIT 1 OFFSET 1) AS SecondHighestSalary;
    View Code
     
     
     
     
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  • 原文地址:https://www.cnblogs.com/pegasus923/p/7654431.html
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